= Solution
Call the coins $1,2,3$ in the order given. Coin 1 beats the constant score of coin 2 exactly when it shows a head, so coin 1 wins with probability $3/5$. Coin 2 beats coin 3 exactly when coin 3 shows a head and scores three, again with probability $3/5$.
Coin 3 beats coin 1 whenever coin 3 shows a tail, or when coin 3 shows a head and coin 1 shows a tail. Its winning probability is therefore
$$
\frac25+\frac35\frac25
=\frac{16}{25}>\frac12.
$$
Thus the preferences form a nontransitive cycle:
$$
1\text{ beats }2,\qquad
2\text{ beats }3,\qquad
3\text{ beats }1.
$$
The second chooser can always select a coin that has winning probability greater than one half against the first choice. Therefore
$$
\boxed{\text{it is preferable to choose second}.}
$$
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